Theorem ddif 4094

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif	⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		velcomp 3919	. . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
2	1	con2bii 360	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
3		vex 3458	. . . 4 ⊢ 𝑥 ∈ V
4	3	biantrur 539	. . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
5	2, 4	bitr2i 279	. 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
6	5	difeqri 4082	1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 400   = wceq 1569   ∈ wcel 2142  Vcvv 3454   ∖ cdif 3901

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456  df-dif 3907

This theorem is used by:  complss  4104  dfun3  4228  dfin3  4229  invdif  4231  ssindif0  4423  difdifdir  4451